Three dice are rolled. What is the probability that the sum of the numbers on…

Concept: For 3 independent dice, each of the 63 ordered outcomes is equally likely, so probability = (favorable ordered outcomes) / (total ordered outcomes). Every unordered combination of face values contributes 3!/(repeat!) ordered arrangements to the favorable count: 6 arrangements when all three values are different, 3 when exactly two match, and 1 when all three match.

Application:

1. Total outcomes: with 3 dice, each face 1 to 6, total ordered outcomes = 6 × 6 × 6 = 216.

2. List every unordered combination of three faces (each between 1 and 6) whose sum is 12: {1,5,6}, {2,4,6}, {2,5,5}, {3,3,6}, {3,4,5}, {4,4,4}.

3. Count the ordered arrangements of each combination using the repeat rule: {1,5,6}, {2,4,6}, {3,4,5} are all-distinct (6 arrangements each); {2,5,5}, {3,3,6} have one repeated pair (3 arrangements each); {4,4,4} has all three equal (1 arrangement).

4. Add the arrangements: 6 + 6 + 6 + 3 + 3 + 1 = 25 favorable outcomes.

5. Probability = favorable outcomes / total outcomes = 25/216.

Cross-check:

Independent check (stars-and-bars): substituting a' = a − 1 (and similarly for b, c) turns a + b + c = 12 with 1 ≤ each ≤ 6 into a' + b' + c' = 9 with 0 ≤ each ≤ 5. The number of non-negative integer solutions without the upper bound is C(11,2) = 55. Subtract the cases where one variable exceeds 5: setting that variable to (value − 6) gives a smaller sum of 3, which has C(5,2) = 10 solutions; this can happen for any of the 3 variables, so subtract 3 × 10 = 30 (two variables cannot both exceed 5 at once, since 6 + 6 already exceeds 9). This gives 55 − 30 = 25, matching the direct enumeration and confirming the probability is 25/216.